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题目:
Every natural number, n
, may have a prime factorization like:
We define the geometric derivative of n, as a number with the following value:
For example: calculate the value of n*
for n = 24500
.
24500 = 2²5³7²
n* = (2*2) * (3*5²) * (2*7) = 4200
Make a function, f
that can perform this calculation
f(n) ----> n*
Every prime number will have n* = 1
.
Every number that does not have an exponent ki
, higher than 1, will give a n* = 1
, too
f(24500) == 4200
f(997) == 1
Do your best!
题意:
题意就是按照那些公式算;首先把一个数分解成质因数相乘;然后按照第二个公式计算即可。
思路:
思路就是先分解成质因数相乘,然后保存在列表里;
然后利用列表的计数函数,计算出每个质因子的个数就是kn;
按照公式求就ok了
第一次提交的代码如下超时了:
import math
from collections import Counter
def f(n):
m=n
nn=1
list1=[]
for i in range(2,m):
while m%i==0:
m=m/i
list1.append(i)
result = Counter(list1)
for key in result:
nn=nn*(result[key]*(key**(result[key]-1)))
if m==n:return 1
return nn
我之前写过类似的知道超时原因是什么? 就是这个m太大会导致超时,解决办法就是处理一下m变成根号m:
import math
from collections import Counter
def f(n):
m=n
nn=1
list1=[]
for i in range(2,int(math.sqrt(m))+1):
while m%i==0:
m=m/i
list1.append(i)
result = Counter(list1)
for key in result:
nn=nn*(result[key]*(key**(result[key]-1)))
if m==n:return 1
return nn
然后就可以了。